SOME INEQUALITIES FOR THE FINITE HILBERT TRANSFORM OF A PRODUCT

S.S. DRAGOMIR

Abstract. Some inequalities for the Hilbert transform of the product of two functions are given.

1. Introduction

Let Ω = (−1, 1) where 1 ≤ p < ∞, the usual 𝔏^{p}−space with respect to the Lebesgue measure λ restricted to the open interval Ω will be denoted by 𝔏^{p} (Ω).

We define a linear operator T (see [24]) from the vector space 𝔏^{1} (Ω) into the vector space of all λ−measurable functions on Ω as follows. Let f ∈ 𝔏^{1} (Ω). The Cauchy principal value

(1.1)  \frac{1}{π} P V ∫_{−1}^{1} \frac{f (τ)}{τ − t}dτ = lim_{ε↓0} [∫_{−1}^{t−ε} + ∫_{t+ε}^{1}] \frac{f (τ)}{π (τ − t)}dτ

exists for λ−almost every t ∈ Ω.

We denote the left-hand side of (1.1) by (T f) (t) for each t ∈ Ω for which (T f) (t) exists. The so-defined function T f, which we call the finite Hilbert Transform of f, is defined λ−almost everywhere on Ω and is λ−measurable; (see for example [1, Theorem 8.1.5]). The resulting linear operator T will be called the finite Hilbert transform operator or Cauchy kernel operator.

It is known that 𝔏^{1} (Ω) is not invariant under T, namely, T (𝔏^{1} (Ω)) ⊄ 𝔏^{1} (Ω) [17, Proof of Theorem 1 (b)].

The following basic results are well known and their proofs may be found in Propositions 8.1.9 and 8.2.1 of [1] respectively. Theorem 1. (M. Riesz) Let 1 < p < ∞. Then T (𝔏^{p} (Ω)) ⊂ 𝔏^{p} (Ω) and the linear operator

T_{p} : f ↦ T f, f ∈ 𝔏^{p} (Ω)

on 𝔏^{p} (Ω) is continuous. Theorem 2. (Parseval) Let 1 < p < ∞ and q = \frac{p}{p−1}. Then

(1.2)  ∫_{−1}^{1} (fT g + gT f) dλ = 0

for every f ∈ 𝔏^{p} (Ω) and g ∈ 𝔏^{q} (Ω).

We introduce the following definition.

Date: May 29, 2001.

1991 Mathematics Subject Classification. Primary 26D15; Secondary 26D10.

Key words and phrases. Hilbert Transform, Analytic Inequalities.

Definition 1. A function f : Ω → ℂ is said to be α−Hölder continuous (0 < α ≤ 1) in a subinterval Ω_{0} of Ω if there exists a constant c > 0, dependent upon Ω_{0}, such that

(1.3)  |f (s) − f (t)| ≤ c |s − t|^{α} , s, t ∈ Ω_{0}.

A function on Ω is said to be locally α−Hölder continuous if it is α−Hölder continuous in every compact subinterval of Ω. We denote by H_{loc}^{α} (Ω) the space of all locally α−Hölder continuous functions on Ω.

The class of Hölder continuous functions on Ω is independent because the finite Hilbert transform of such a function exists everywhere on Ω (see [15, Section 3.2] or [21, Lemma II.1.1]).

This is in contrast to the λ−almost everywhere existence of the finite Hilbert transform of functions in 𝔏^{1} (Ω).

There are continuous functions f ∈ 𝔏^{1} (Ω) such that (T f) (t) does not exist at some point t ∈ Ω. An example is given by the function f defined by (see [24])

f (t) = \{ \begin{matrix}0 & if & −1 < t ≤ 0, \\ \frac{1}{ln t−ln 2} & if & 0 < t < 1. \end{matrix}

It readily follows that (T f) (0) does not exist.

In paper [24] it is proved amongst others the following result. Theorem 3. (Okada-Elliot) The space 𝔏^{p} (Ω) ∩ H_{loc}^{α} (Ω) is invariant under the finite Hilbert transform operator T and the restriction of T to that space is continuous whenever 1 < p < ∞. This, however, is not true when p = 1.

We consider the finite Hilbert transform on the open interval (a, b)

(T f) (a, b; t) := \frac{1}{π} P V ∫_{a}^{b} \frac{f (τ)}{τ − t}dτ, t ∈ (a, b) .

The following theorem holds (see [11]). Theorem 4. Let f : [a, b] → ℝ be α − H−Hölder continuous on (a, b), i.e.,

(1.4)  |f (t) − f (s)| ≤ H |t − s|^{α} for all t, s ∈ (a, b) , α ∈ (0, 1], H > 0.

Then we have the estimate

|(T f) (a, b; t) − \frac{f (t)}{π} ln (\frac{b − t}{t − a})| (1.5)

≤ \frac{H}{απ} [(t − a)^{α} + (b − t)^{α}] ≤ \frac{H2^{1−α}}{απ} (b − a)^{α} ,

for all t ∈ (a, b).

The following result holds for monotonic functions (see [11]). Theorem 5. Let f : [a, b] → ℝ be a monotonic nondecreasing (nonincreasing) function on [a, b]. If the finite Hilbert transform (T f) (a, b, ·) exists in every t ∈ (a, b), then

(1.6)  (T f) (a, b; t) ≥ (≤) \frac{1}{π} f (t) ln (\frac{b − t}{t − a})

for all t ∈ (a, b).

Now, if we assume that the mapping f : (a, b) → ℝ is convex on (a, b), then it is locally Lipschitzian on (a, b) and then the finite Hilbert transform of f exists in every point t ∈ (a, b).

The following result holds (see [11]). Theorem 6. Let f : (a, b) → ℝ be a convex mapping on (a, b). Then we have

\frac{1}{π} [l (t) (b − t) + ∫_{a}^{t} l (s) ds + f (t) ln (\frac{b − t}{t − a})] (1.7)

≤ (T f) (a, b; t)
≤ \frac{1}{π} [f (t) ln (\frac{b − t}{t − a}) + l (t) (t − a) + ∫_{t}^{b} l (s) ds] ,

where l (s) ∈ [f′_{−} (s) , f′_{+} (s)], s ∈ (a, b).

The following more practical result also holds [11]: Corollary 1. Let f : (a, b) → ℝ be a differentiable convex function on (a, b). Then we have the inequality

\frac{1}{π} [f (t) − f (a) + f′ (t) (b − t) + f (t) ln (\frac{b − t}{t − a})] (1.8)

≤ (T f) (a, b; t)
≤ \frac{1}{π} [f (t) ln (\frac{b − t}{t − a}) + f (b) − f (t) + f′ (t) (t − a)]

for all t ∈ (a, b).

In this paper we point out some inequalities for the finite Hilbert transform of the product of two functions.

For a comprehensive number of results on the numerical approximation of the Cauchy principal value integrals, see [2] – [10], [13] – [14], [16], [18] – [20], [22] – [23], [25] – [27].

2. The Results

The following lemma holds. Lemma 1. If f and g are locally Hölder continuous on [a, b] , then fg is also locally Hölder continuous on [a, b] and:

T (fg) (a, b; t) (2.1)

= f (t) T (g) (a, b; t) + g (t) T (f) (a, b; t)

−\frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a}) + \frac{1}{π} P V ∫_{a}^{b} \frac{(f (τ) − f (t)) (g (τ) − g (t))}{τ − t}dτ

for any t ∈ (a, b) .

Proof. Assume that for a subinterval [c, d] ⊆ [a, b] , we have

(2.2)  |f (s) − f (u)| ≤ L_{1} |s − u|^{r_{1}} for any s, u ∈ [c, d] ;

(2.3)  |g (s) − g (u)| ≤ L_{2} |s − u|^{r_{2}} for any s, u ∈ [c, d] .

Then

|f (s) g (s) − f (u) g (u)| = |f (s) g (s) − f (s) g (u) + f (s) g (u) − f (u) g (u)|
≤ |f (s)| |g (s) − g (u)| + |g (u)| |f (s) − f (u)|
≤ M_{1}L_{1} |s − u|^{r_{1}} + M_{2}L_{2} |s − u|^{r_{2}}
≤ |s − u|^{r} [M_{1}L_{1} |s − u|^{r_{1}−r} + M_{2}L_{2} |s − u|^{r_{2}−r}]
≤ |s − u|^{r} [M_{1}L_{1} |d − c|^{r_{1}−r} + M_{2}L_{2} |d − c|^{r_{2}−r}]
= M |s − u|^{r}

where

M_{1} := sup_{s∈[c,d]} |f (s)| , M_{2} := sup_{u∈[c,d]} |g (u)| , r = min (r_{1}, r_{2}) ,

and

M = M_{1}L_{1} |d − c|^{r_{1}−r} + M_{2}L_{2} |d − c|^{r_{2}−r} ,

proving that fg is locally Hölder continuous on [a, b] . Now, for any t, τ ∈ [a, b] , we may write that

(f (τ) − f (t)) (g (τ) − g (t)) = f (τ) g (τ) + f (t) g (t) − f (t) g (τ) − f (τ) g (t)

giving

\frac{f (τ) g (τ)}{τ − t} = f (t) · \frac{g (τ)}{τ − t} + g (t) · \frac{f (τ)}{τ − t} − \frac{f (t) g (t)}{τ − t}

+\frac{(f (τ) − f (t)) (g (τ) − g (t))}{τ − t}

for any t, τ ∈ [a, b] , t ≠ τ.

Consequently,

T (fg) (a, b; t)

= \frac{1}{π} P V ∫_{a}^{b} \frac{f (τ) g (τ)}{τ − t}dτ
= \frac{1}{π} f (t) P V ∫_{a}^{b} \frac{g (τ)}{τ − t}dτ + g (t) \frac{1}{π} P V ∫_{a}^{b} ·\frac{f (τ)}{τ − t}dτ

−\frac{1}{π} f (t) g (t) P V ∫_{a}^{b} \frac{dτ}{τ − t} + \frac{1}{π} P V ∫_{a}^{b} \frac{(f (τ) − f (t)) (g (τ) − g (t))}{τ − t}dτ

= f (t) T (g) (a, b; t) + g (t) T (f) (a, b; t)

−\frac{f (t) g (t)}{π} ln (\frac{b − t}{t − a}) + \frac{1}{π} P V ∫_{a}^{b} \frac{(f (τ) − f (t)) (g (τ) − g (t))}{τ − t}dτ

for any t ∈ (a, b) , and the identity (2.1) is proved.

Theorem 7. Assume that f is of L_{1} − r_{1}−Hölder type and g is of L_{2} − r_{2}−Hölder type on [a, b] , where L_{1}, L_{2} > 0, r_{1}, r_{2} ∈ (0, 1]. Then we have the inequality:

|T (fg) (a, b; t) − f (t) T (g) (a, b; t) (2.4)

− g (t) T (f) (a, b; t) + \frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a})|

≤ \frac{L_{1}L_{2}}{π (r_{1} + r_{2})} [(b − t)^{r_{1}+r_{2}} + (t − a)^{r_{1}+r_{2}}]
≤ \frac{L_{1}L_{2} (b − a)^{r_{1}+r_{2}}}{π (r_{1} + r_{2})}

for any t ∈ (a, b) .

Proof. Taking the modulus in (2.1), we may write

|T (fg) (a, b; t) − f (t) T (g) (a, b; t)

−g (t) T (f) (a, b; t) + \frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a})|

≤ \frac{1}{π} P V ∫_{a}^{b} |\frac{(f (τ) − f (t)) (g (τ) − g (t))}{τ − t}| dτ ≤ \frac{1}{π} P V ∫_{a}^{b} L_{1}L_{2} |τ − t|^{r_{1}+r_{2}−1} dτ

= \frac{L_{1}L_{2}}{π} [\frac{(b − t)^{r_{1}+r_{2}} + (t − a)^{r_{1}+r_{2}}}{r_{1} + r_{2}} ]

and the first part of inequality (2.4) is proved. The second part is obvious.

The best inequality we can get from (2.4) is embodied in the following corollary.

Corollary 2. With the assumptions in Theorem 7, we have

|T (fg) (a, b; \frac{a + b}{2}) − f (\frac{a + b}{2}) T (g) (a, b; \frac{a + b}{2}) (2.5)

− g (\frac{a + b}{2}) T (f) (a, b; \frac{a + b}{2})|

≤ \frac{L_{1}L_{2} (b − a)^{r_{1}+r_{2}}}{π (r_{1} + r_{2}) 2^{r_{1}+r_{2}−1}} .

The following corollary also holds.

Corollary 3. If f and g are Lipschitzian with the constants K_{1} and K_{2}, then we have the inequality (2.6)

|T (fg) (a, b; t) − f (t) T (g) (a, b; t) − g (t) T (f) (a, b; t) + \frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a})|

(2.7)  ≤ \frac{K_{1}K_{2}}{π} [\frac{1}{4} (b − a)^{2} + (t − \frac{a + b}{2})^{2}] ≤ \frac{K_{1}K_{2}}{2π} (b − a)^{2}

for any t ∈ (a, b) . In particular, for t = \frac{a+b}{2}, we have (2.8)

|T (fg) (a, b; \frac{a + b}{2}) − f (\frac{a + b}{2}) T (g) (a, b; \frac{a + b}{2})−g (\frac{a + b}{2}) T (f) (a, b; \frac{a + b}{2})|

(2.9)  ≤ \frac{K_{1}K_{2}}{4π} (b − a)^{2} .

The following theorem also holds. Theorem 8. Assume that f and g are absolutely continuous on [a, b] . Then we have the inequality:

|T (fg) (a, b; t) − f (t) T (g) (a, b; t) (2.10)

− g (t) T (f) (a, b; t) + \frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a})|

(2.11)

≤ \frac{1}{π} ×\{ \begin{matrix}[\frac{1}{4} (b − a)^{2} + (t − \frac{a + b}{2})^{2}] & ∥f′∥ & [a,b],∞ & ∥g′∥ & [a,b],∞ \\ if f′ ∈ L_{∞} [a, b] , g′ ∈ L_{∞} [a, b] ; \\ \frac{δ}{δ + 1} & [(b − t)^{1+ \frac{1}{δ}} + (t − a)^{1+ \frac{1}{δ}} ] & ∥f′∥ & [a,b],∞ & ∥g′∥ & [a,b],γ \\ if f′ ∈ L_{∞} [a, b] , g′ ∈ L_{γ} [a, b] , γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (b − a) ∥f′∥_{[a,b],∞} ∥g′∥_{[a,b],1}  if f′ ∈ L_{∞} [a, b] , g′ ∈ L_{1} [a, b] ; \\ \frac{β}{β + 1} & [(b − t)^{1+ \frac{1}{β}} + (t − a)^{1+ \frac{1}{β}} ] & ∥f′∥ & [a,b],α & ∥g′∥ & [a,b],∞ \\ if f′ ∈ L_{α} [a, b] , α > 1, \frac{1}{α} + \frac{1}{β} = 1, and g′ ∈ L_{∞} [a, b] ; \\ \frac{βδ}{β + δ} & [(b − t)^{\frac{δ+β}{β+δ}} + (t − a)^{\frac{δ+β}{β+δ}} ] & ∥f′∥ & [a,b],α & ∥g′∥ & [a,b],γ \\ if f′ ∈ L_{α} [a, b] , α > 1, \frac{1}{α} + \frac{1}{β} = 1, and g′ ∈ L_{γ} [a, b] , γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ β & [(b − t)^{\frac{1}{β}} + (t − a)^{\frac{1}{β}} ] & ∥f′∥ & [a,b],α & ∥g′∥ & [a,b],1 \\ if f′ ∈ L_{α} [a, b] , α > 1, \frac{1}{α} + \frac{1}{β} = 1, and g′ ∈ L_{1} [a, b] ; \\ (b − a) ∥f′∥_{[a,b],1} ∥g′∥_{[a,b],∞}  if f′ ∈ L_{1} [a, b] , g′ ∈ L_{∞} [a, b] ; \\ δ & [(b − t)^{1+ \frac{1}{δ}} + (t − a)^{1+ \frac{1}{δ}} ] & ∥f′∥ & [a,b],1 & ∥g′∥ & [a,b],γ \\ if f′ ∈ L_{1} [a, b] , g′ ∈ L_{γ} [a, b] , γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \end{matrix}

Proof. Since f and g are absolutely continuous on [a, b] , we may write that

f (τ) − f (t) = ∫_{t}^{τ} f′ (u) du and g (τ) − g (t) = ∫_{t}^{τ} g′ (u) du

which implies:

(2.12)  |f (τ) − f (t)| ≤ \{ \begin{matrix}∥f′∥_{[τ,t],∞} |τ − t| & if f′ ∈ L_{∞} [a, b] ; \\ ∥f′∥_{[τ,t],α} |τ − t|^{\frac{1}{β}} & if f′ ∈ L_{α} [a, b] , \\ & α > 1, \frac{1}{α} + \frac{1}{β} = 1, \\ ∥f′∥ & [τ,t],1 \end{matrix}

and

(2.13)  |g (τ) − g (t)| ≤ \{ \begin{matrix}∥g′∥_{[τ,t],∞} |τ − t| & if g′ ∈ L_{∞} [a, b] ; \\ ∥g′∥_{[τ,t],γ} |τ − t|^{\frac{1}{δ}} & if g′ ∈ L_{γ} [a, b] , \\ & γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ ∥g′∥_{[τ,t],1} . \end{matrix}

Using the identity (2.2), we get

|T (fg) (a, b; t) − f (t) T (g) (a, b; t) (2.14)

−g (t) T (f) (a, b; t) + \frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a}) |

≤ \frac{1}{π} P V ∫_{a}^{b} |\frac{(f (τ) − f (t)) (g (τ) − g (t))}{τ − t}| dτ =: I.

Then we have, by using (2.12) or (2.13), that

(2.15)  I ≤ \frac{1}{π} × \{ \begin{matrix}P V ∫ _{a}^{b} ∥f′∥_{[τ,t],∞} ∥g′∥_{[τ,t],∞} |τ − t| dτ \\ P V ∫ _{a}^{b} ∥f′∥_{[τ,t],∞} ∥g′∥_{[τ,t],α} |τ − t|^{\frac{1}{δ}} dτ \\ P V ∫ _{a}^{b} ∥f′∥_{[τ,t],∞} ∥g′∥_{[τ,t],1} dτ \\ P V ∫ _{a}^{b} ∥f′∥_{[τ,t],α} ∥g′∥_{[τ,t],∞} |τ − t|^{\frac{1}{β}} dτ \\ P V ∫ _{a}^{b} ∥f′∥_{[τ,t],α} ∥g′∥_{[τ,t],γ} |τ − t|^{\frac{1}{β} + \frac{1}{δ} −1} dτ \\ P V ∫ _{a}^{b} ∥f′∥_{[τ,t],α} ∥g′∥_{[τ,t],1} |τ − t|^{\frac{1}{β} −1} dτ \\ P V ∫ _{a}^{b} ∥f′∥_{[τ,t],1} ∥g′∥_{[τ,t],∞} dτ \\ P V ∫ _{a}^{b} ∥f′∥_{[τ,t],1} ∥g′∥_{[τ,t],γ} |τ − t|^{\frac{1}{δ} −1} dτ \\ P V ∫ _{a}^{b} ∥f′∥_{[τ,t],1} ∥g′∥_{[τ,t],1} |τ − t|^{−1} dτ. \end{matrix}

However,

P V ∫_{a}^{b} ∥f′∥_{[τ,t],∞} ∥g′∥_{[τ,t],∞} |τ − t| dτ

≤ ∥f′∥_{[a,b],∞} ∥g′∥_{[a,b],∞} [\frac{(b − t)^{2} + (t − a)^{2}}{2} ]
= ∥f′∥_{[a,b],∞} ∥g′∥_{[a,b],∞} [\frac{1}{4} (b − a)^{2} + (t − \frac{a + b}{2})^{2}] ,

P V ∫_{a}^{b} ∥f′∥_{[τ,t],∞} ∥g′∥_{[τ,t],α} |τ − t|^{\frac{1}{δ}} dτ

≤ ∥f′∥_{[a,b],∞} ∥g′∥_{[a,b],γ} [\frac{(b − t)^{1+ \frac{1}{δ}} + (t − a)^{1+ \frac{1}{δ}}}{\frac{1}{δ} + 1} ]
= \frac{δ}{δ + 1} ∥f′∥_{[a,b],∞} ∥g′∥_{[a,b],γ} [(b − t)^{1+ \frac{1}{δ}} + (t − a)^{1+ \frac{1}{δ}} ] ,

P V ∫_{a}^{b} ∥f′∥_{[τ,t],∞} ∥g′∥_{[τ,t],1} dτ ≤ ∥f′∥_{[a,b],∞} ∥g′∥_{[a,b],1} (b − a) ,

P V ∫_{a}^{b} ∥f′∥_{[τ,t],α} ∥g′∥_{[τ,t],∞} |τ − t|^{\frac{1}{β}} dτ

≤ ∥f′∥_{[a,b],α} ∥g′∥_{[a,b],∞} · \frac{β}{β + 1} [(b − t)^{\frac{1}{β} +1} + (t − a)^{\frac{1}{β} +1}] ,

P V ∫_{a}^{b} ∥f′∥_{[τ,t],α} ∥g′∥_{[τ,t],γ} |τ − t|^{\frac{1}{β} + \frac{1}{δ} −1} dτ

≤ ∥f′∥_{[a,b],α} ∥g′∥_{[a,b],γ} \frac{1}{\frac{1}{β} + \frac{1}{δ}} [(b − t)^{\frac{1}{β} + \frac{1}{δ}} + (t − a)^{\frac{1}{β} + \frac{1}{δ}} ]
= \frac{βδ}{β + δ} ∥f′∥_{[a,b],α} ∥g′∥_{[a,b],γ} [(b − t)^{\frac{δ+β}{β+δ}} + (t − a)^{\frac{δ+β}{β+δ}} ] ,

P V ∫_{a}^{b} ∥f′∥_{[τ,t],α} ∥g′∥_{[τ,t],1} |τ − t|^{\frac{1}{β} −1} dτ

≤ ∥f′∥_{[a,b],α} ∥g′∥_{[a,b],1} β [(b − t)^{\frac{1}{β}} + (t − a)^{\frac{1}{β}} ] ,

P V ∫_{a}^{b} ∥f′∥_{[τ,t],1} ∥g′∥_{[τ,t],∞} dτ ≤ (b − a) ∥f′∥_{[a,b],1} ∥g′∥_{[a,b],∞}

and

P V ∫_{a}^{b} ∥f′∥_{[τ,t],1} ∥g′∥_{[τ,t],γ} |τ − t|^{\frac{1}{δ} −1} dτ

≤ ∥f′∥_{[a,b],1} ∥g′∥_{[a,b],γ} δ [(b − t)^{1+ \frac{1}{δ}} + (t − a)^{1+ \frac{1}{δ}} ] .

For the last inequality we cannot point out a bound as above. Using (2.14) and (2.15), we deduce the desired inequality (2.10).

The following lemma also holds.

Lemma 2. Let f : [a, b] → ℝ be locally Hölder continuous on [a, b] and g : [a, b] → ℝ so that g′ is absolutely continuous on [a, b]. Then we have the identity:

T (fg) (a, b; t) (2.16)

= f (t) T (g) (a, b; t) + g (t) T (f) (a, b; t) − \frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a})

+\frac{1}{π} [∫_{a}^{b} f (τ) dτ − (b − a) f (t)] g′ (t)

−\frac{1}{π} P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t} (∫_{t}^{τ} (u − τ) g′′ (u) du) dτ

for any t ∈ (a, b) .

Proof. We use the following identity:

∫_{α}^{β} φ (u) du = φ (α) (β − α) − ∫_{α}^{β} (u − β) φ′ (u) du

which holds for any absolutely continuous function φ : [α, β] → ℝ.

Then we have

\frac{1}{π} P V ∫_{a}^{b} [f (τ) − f (t)] [\frac{1}{τ − t} (g (τ) − g (t))] dτ

= \frac{1}{π} P V ∫_{a}^{b} [f (τ) − f (t)] [\frac{1}{τ − t} ∫_{t}^{τ} g′ (u) du] dτ

= \frac{1}{π} P V ∫_{a}^{b} [f (τ) − f (t)] [g′ (t) − \frac{1}{τ − t} ∫_{t}^{τ} (u − τ) g′′ (u) du] dτ

= \frac{1}{π} [g′ (t) ∫_{a}^{b} f (τ) dτ − (b − a) f (t) g′ (t)

− P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t} (∫_{t}^{τ} (u − τ) g′′ (u) du) dτ]

= \frac{1}{π} g′ (t) ∫_{a}^{b} f (τ) dτ − \frac{1}{π} (b − a) f (t) g′ (t)

−\frac{1}{π} P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t} (∫_{t}^{τ} (u − τ) g′′ (u) du) dτ.

Using (2.1), we deduce (2.16).

The following theorem holds.

Theorem 9. Assume that f : [a, b] → ℝ is of H − r−Hölder type and g : [a, b] → ℝ is such that g′ is absolutely continuous on [a, b] . Then we have the inequality:

|T (fg) (a, b; t) − f (t) T (g) (a, b; t) − g (t) T (f) (a, b; t) (2.17)

+\frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a}) − \frac{1}{π} [∫_{a}^{b} f (τ) dτ − (b − a) f (t)] g′ (t)|

≤ \frac{H}{π} \{ \begin{matrix}\frac{1}{2(r+2)} [(b − t)^{r+2} + (t − a)^{r+2}] ∥g′′∥_{[a,b],∞} & if g′′ ∈ L_{∞} [a, b] ; \\ _{\frac{q}{(rq+q+1)(q+1)^{\frac{1}{q}}}} [(b − t)^{r+ \frac{1}{q} +1} + (t − a)^{r+ \frac{1}{q} +1}] ∥g′′∥_{[a,b],p} & if g′′ ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{(r+1)} [(b − t)^{r+1} + (t − a)^{r+1}] ∥g′′∥_{[a,b],1} . \end{matrix}

Proof. Using the identity (2.16), we deduce that the left side in (2.17) is upper bounded by

I : = \frac{1}{π} P V ∫_{a}^{b} |f (τ) − f (t)| \frac{1}{|τ − t|} |∫_{t}^{τ} (u − τ) g′′ (u) du| dτ

≤ \frac{H}{π} P V ∫_{a}^{b} |τ − t|^{r−1} |∫_{t}^{τ} (u − τ) g′′ (u) du| dτ =: J.

We observe that

|∫_{t}^{τ} (u − τ) g′′ (u) du| ≤ ∥g′′∥_{[t,τ],∞} \frac{(τ − t)^{2}}{2}

if g′′ ∈ L_{∞} [a, b] ,

|∫_{t}^{τ} (u − τ) g′′ (u) du| ≤ ∥g′′∥_{[t,τ],p} |∫_{t}^{τ} |t − τ|^{q} dτ|^{\frac{1}{q}} = ∥g′′∥_{[t,τ],p} \begin{matrix}|t − τ| & \frac{q+1}{q} \\ (q + 1) & \frac{1}{q} \end{matrix}

if g′′ ∈ L_{p} [a, b] and, finally,

|∫_{t}^{τ} (u − τ) g′′ (u) du| ≤ |t − τ| ∥g′′∥_{[t,τ],1} .

Consequently, we have

J ≤ \frac{H}{π} × \{ \begin{matrix}P V ∫ _{a}^{b} |τ − t|^{r−1} · \frac{(τ−t)^{2}}{2} · ∥g′′∥_{[t,τ],∞} dτ \\ P V ∫ _{a}^{b} \frac{|τ − t|^{r−1} · |t − τ|^{1+ \frac{1}{q}}}{(q + 1)^{\frac{1}{q}}} ∥g′′∥_{[t,τ],p} dτ \\ P V ∫ _{a}^{b} |τ − t|^{r−1} · |t − τ| ∥g′′∥_{[t,τ],1} dτ \end{matrix}

≤ \frac{H}{π} × \{ \begin{matrix}\frac{1}{2} & ∥g′′∥ & [a,b],∞ & [\frac{(b − t)^{r+2} + (t − a)^{r+2}}{r + 2} ] \\ \frac{1}{(q + 1)^{\frac{1}{q}}} & ∥g′′∥ & [a,b],p & [\frac{(b − t)^{r+ \frac{1}{q} +1} + (t − a)^{r+ \frac{1}{q} +1}}{r + \frac{1}{q} + 1} ] \\ ∥g′′∥_{[a,b],1} · [\frac{(b − t)^{r+1} + (t − a)^{r+1}}{r + 1} ] , \end{matrix}

which proves the inequality (2.17).

The following lemma also holds. Lemma 3. Assume that f and g are as in Lemma 2. Then we have the identity: T (fg) (a, b; t) (2.18)

= f (t) T (g) (a, b; t) + g (t) T (f) (a, b; t) − \frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a})

+\frac{1}{π} [∫_{a}^{b} f (τ) g′ (τ) dτ − [g (b) − g (a)] f (t)]

−\frac{1}{π} P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t} (∫_{t}^{τ} (u − t) g′′ (u) du) dτ

for any t ∈ (a, b) . Proof. In this case, we use the following identity:

∫_{α}^{β} φ (u) du = φ (β) (β − α) − ∫_{α}^{β} (u − α) φ′ (u) du

which holds for any absolutely continuous function φ : [α, β] → ℝ.

FINITE HILBERT TRANSFORM 13

Then, as above, we have

\frac{1}{π} P V ∫_{a}^{b} [f (τ) − f (t)] [\frac{1}{τ − t} (g (τ) − g (t))] dτ

= \frac{1}{π} P V ∫_{a}^{b} [f (τ) − f (t)] [\frac{1}{τ − t} ∫_{t}^{τ} g′ (u) du] dτ
= \frac{1}{π} P V ∫_{a}^{b} [f (τ) − f (t)] [g′ (t) − \frac{1}{τ − t} ∫_{t}^{τ} (u − τ) g′′ (u) du] dτ
= \frac{1}{π} [∫_{a}^{b} f (τ) g′ (τ) dτ − f (t) P V ∫_{a}^{b} g′ (τ) dτ

− P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t} (∫_{t}^{τ} (u − t) g′′ (u) du) dτ]

= \frac{1}{π} [∫_{a}^{b} f (τ) g′ (τ) dτ − [g (b) − g (a)] f (t)

− P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t} (∫_{t}^{τ} (u − t) g′′ (u) du) dτ] ,

proving the identity (2.18).

The following result also holds. Theorem 10. With the assumptions in Theorem 9, we have: |T (fg) (a, b; t) − f (t) T (g) (a, b; t) − g (t) T (f) (a, b; t) (2.19)

+\frac{1}{π} f (t) g (t) ln (\frac{b − t}{t − a}) − \frac{1}{π} [∫_{a}^{b} f (τ) g′ (τ) dτ − [g (b) − g (a)] f (t)]|

≤ \frac{H}{π} \{ \begin{matrix}\frac{1}{2(r+2)} [(b − t)^{r+2} + (t − a)^{r+2}] ∥g′′∥_{[a,b],∞} & if g′′ ∈ L_{∞} [a, b] ; \\ _{\frac{q}{(rq+q+1)(q+1)^{\frac{1}{q}}}} [(b − t)^{r+ \frac{1}{q} +1} + (t − a)^{r+ \frac{1}{q} +1}] ∥g′′∥_{[a,b],p} & if g′′ ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{(r+1)} [(b − t)^{r+1} + (t − a)^{r+1}] ∥g′′∥_{[a,b],1} . \end{matrix}

Proof. The proof follows in a similar manner to the one in Theorem 9 by the use of Lemma 3. We omit the details.

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